Solution:
Let $ABCD$ be a tetrahedron of side $s$. We want to find the distance between two of its opposite sides. Let $E$ be the midpoint of $AD$, $F$ the midpoint of $BC$. Then $AE = s/2$, $AF = s\sqrt{3}/2$, and $\angle AEF = 90^{\circ}$. So the distance between the two opposite sides is $EF = \sqrt{AF^2 - AE^2} = \sqrt{3s^2/4 - s^2/4} = s/\sqrt{2}$.

Now we find the distance between a body diagonal and a face diagonal of a cube of side $a$. Let $O$ be the center of the cube and $P$ be the midpoint of the face diagonal. Then the plane containing $P$ and the body diagonal is perpendicular to the face diagonal. So the distance between the body and face diagonals is the distance between $P$ and the body diagonal, which is $\frac{a}{2} \sqrt{\frac{2}{3}}$ (the altitude from $P$ of right triangle $OPQ$, where $Q$ is the appropriate vertex of the cube). So now $\frac{s}{\sqrt{2}} = \frac{a}{2} \sqrt{\frac{2}{3}}$, thus $s = a/\sqrt{3} = 12/\sqrt{3} = 4\sqrt{3}$.